Atoms of metals x, y and z form face-centred cubic (fcc) unit cell...

Metal $x$ forms FCC (edge length ${L}_{x}$ )

Metal y forms BCC (edge length ${L}_{y}$ )

Metal $z$ forms SC (edge length ${L}_{z}$ )

Given ${r}_{z} = \dfrac{\sqrt{3}}{2}{r}_{y}$ and ${r}_{y} = \dfrac{8}{\sqrt{3}}{r}_{x}$

$∴ {r}_{z} = \dfrac{\sqrt{3}}{2} \times \dfrac{8}{\sqrt{3}}{r}_{x} = 4{r}_{x}$

${M}_{z} = \dfrac{3}{2}{M}_{y}{M}_{z} = 3{M}_{x}$

$∴ {M}_{y} = 2{M}_{x}$

Packing efficiency $FCC > BCC > SC$

Packing efficiency unit cell $x > y > z$

In FCC unit cell:- atoms along the face diagonals are in contact.

$∴ \sqrt{2}{L}_{x} = 4{r}_{x} ⇒ {L}_{x} = 2\sqrt{2}{r}_{x}$

In BCC unit cell: atoms along the body diagonal are

$∴ \sqrt{3}{L}_{y} = 4{r}_{y} ⇒ {L}_{y} = \dfrac{4}{\sqrt{3}}{r}_{y} = \dfrac{4}{\sqrt{3}} \times \dfrac{8}{\sqrt{3}}{r}_{x} = \dfrac{32}{3}{r}_{x}$

${L}_{y} = \dfrac{32}{3}{r}_{x}$

In SC unit cell, atoms along the edge are in contact

$∴ {L}_{z} = 2{r}_{z}$

$= 2 \times 4{r}_{x} = 8{r}_{x}$

${L}_{x} = 2\sqrt{2}{r}_{x}$

${L}_{y} = \dfrac{32}{3}{r}_{x}$

${L}_{z} = 8{r}_{x}$

$∴ {L}_{y} > {L}_{z} > {L}_{x}$

Density of $x$ (Number of atoms of $x$ per unit cell $( z ) = 4 )$

${d}_{x} = \dfrac{{zM}_{x}}{{\left({L}_{x}\right)}^{3}{N}_{A}} = \dfrac{4 \times {M}_{x}}{{\left(2\sqrt{2}{r}_{x}\right)}^{3} \times {N}_{A}}$

$= \dfrac{4{M}_{x}}{16\sqrt{2}{r}_{x}^{3}{N}_{A}} = \dfrac{{M}_{x}}{4\sqrt{2}{r}_{x}^{3}{N}_{A}}$

Density of $y$ : (Number of atoms of y per unit cell $( z ) = 2$ )

${d}_{y} = \dfrac{{zM}_{y}}{{\left({L}_{y}\right)}^{3}{N}_{A}} = \dfrac{2 \times 2{M}_{x}}{{\left(\dfrac{32}{3}{r}_{x}\right)}^{3}{N}_{A}} = \dfrac{108{M}_{x}}{32768{r}_{x}^{3}{N}_{A}}$

$∴$ Density of $x >$ density of $y$.